Two particles each of mass \(m\) are connected by an (inextensible) string of length \(l\). One particle moves on a horizontal table (assume no friction). 'The string passes through a hole in the table and the particle at the lower end moves up and down along a vertical line. Find the Iagrange equations of motion of the particles. Hint: Let the coordinates of the particle on the table be \(r\) and \(\theta\), and let the coordinate of the other particle be \(z\). Eliminate one variable from \(L\) (using \(r+|z|=l)\) and write two Lagrange equations.

Kinetic energy is the energy that an object possesses due to its motion. In classical mechanics, for a particle of mass \(m\) and velocity \(\textbf{v}\), it is given by \(\frac{1}{2}mv^2\). In our problem, the kinetic energy is calculated for two particles. One particle moves on a horizontal table with coordinates \(r\) and \(\theta\), contributing \(\frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2)\) to the kinetic energy. The other particle moves vertically with coordinate \(z\), contributing \( \frac{1}{2}m \dot{z}^2\). Therefore, the total kinetic energy of the system is: \[ T = \frac{1}{2}m( \dot{r}^2 + r^2 \dot{\theta}^2) + \frac{1}{2}m \dot{z}^2 \] This accounts for the motion of both particles, combining their individual kinetic energies.

Potential energy is the energy stored in an object due to its position or configuration. In our exercise, the particle moving vertically along coordinate \(z\) possesses potential energy due to gravity. Setting the zero of potential energy at the table level, the potential energy of the vertically moving particle is \V = mgz\, where \[ g \] is the acceleration due to gravity. This expression shows how the potential energy depends on the height \(z\) of the particle. Here, we only consider the potential energy of the vertically moving particle because the particle on the table experiences no change in potential energy as it moves horizontally.

The Euler-Lagrange equations form the core of the Lagrangian mechanics approach. These equations derive the motion of a system by considering the Lagrangian \(L\), which is the difference between kinetic and potential energies: \[ L = T - V = \frac{1}{2}m(\dot{r}^2 + r^2 \dot{\theta}^2) + \frac{1}{2}m \dot{z}^2 - mgz \] We apply the Euler-Lagrange equation: \frac{d}{dt} \left( \frac{\partial L}{\partial \dot{q}} \right) - \frac{\partial L}{\partial q} = 0\ for each generalized coordinate \(q\). In this problem, these coordinates are \(r\) and \(\theta\). For \r: \[ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{r}} \right) - \frac{\partial L}{\partial r} = 0 \] For \theta: \[ \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{\theta}} \right) - \frac{\partial L}{\partial \theta} = 0 \] Solving these equations provides the equations of motion for the system.

Constraint equations are vital for handling systems where movements are restricted. In our exercise, the string's inextensible nature provides a constraint: \r + |z| = l\. This constraint ensures that the total length of the string remains constant. Such constraints reduce the number of independent variables, simplifying the problem. Here, the constraint allows elimination of \(z\), expressing it as \(z = l - r\). This expression relates the coordinates of both particles and streamlines the process of deriving the Lagrangian and subsequent equations of motion.

Classical mechanics is the branch of physics that studies the motion of bodies under the influence of forces. The Lagrangian formulation, used in this exercise, is a powerful technique within classical mechanics. It focuses on energy concepts—kinetic and potential—making it adaptable to various systems. The Euler-Lagrange equations, derived from the principle of least action, provide a systematic way to find the equations of motion for complex systems with constraints. This exercise, involving two interconnected particles, exemplifies how classical mechanics handles intricate relationships between dynamic variables to predict the system's behavior.